AM 105b J. R. Rice (January 2008 revision of February 1997 notes) Solutions of the Diffusion Equation (a Partial Differential Equation)

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1 AM 15b J. R. Rice (January 28 revision of February 1997 notes) Solutions of the Diffusion Equation (a Partial Differential Equation) Ordinary differential equations (ode's) are important on their own. We also study them because they arise in solving partial differential equations (pde's). These notes show examples of how ode's arise in developing solutions of a particular pde, the diffusion equation (sometimes called the heat equation). That pde in its simplest form requires that a function = (x,t) satisfy 2 x 2 = t (where is a positive constant) Typically x would denote distance and t time, and would then have units of (length) 2 /time. Appendix 1 explains situations leading to that pde, or to its generalization 2 = / t in 2 or 3 space dimensions. E.g., in solution chemistry, could be local concentration of a solute; in thermal physics, could be local temperature in a solid; in economic modeling, could be number of coins of a certain type per unit land area. In studying one approach to solving the pde, by separation of variables as in the next section, we'll also have an accessible introduction to something that we'll study in a much wider context later in the course. That involves generating orthogonal basis functions in terms of which we can develop representations of arbitrary functions as infinite series. In the case we look at here, our orthogonal basis functions are just sine functions, and the representation of arbitrary functions as infinite series of them is just a special form of the well known Fourier series. Don't worry at this point if it seems hard to understand the motivation for some of the steps in developing solutions of the pde. We'll look at such matters more systematically later in the semester. Just view this as an insight into how ode's may arise in solving pde's and as a perhaps unexpected route to the Fourier series. Solution of the diffusion equation for a finite-thickness slab by separation of variables: The problem: Determine = (x,t) which solves the pde 2 x 2 = t in the domain < x < and < t < of the x,t plane subject to the and the initial condition at t=: (x,)= init (x) (some given function) for < x <, boundary conditions at x= and x=: (,t) = and (,t) = for < t <. 1

2 ere is the idea of separation of variables: We begin by seeking expressions for in the simple product form, (x, t) = f (x)g(t), such that we solve the pde and, also, meet the boundary conditions. But we do not yet try to meet the initial conditions. Then, after finding a family of such product terms, we seek a way of summing together an infinite number of that family in a way that does meet the initial conditions. To get started, first note that to solve the pde 2 x = in such product form 2 t (x, t) = f (x)g(t), one must have d2 f (x) dx 2 g(t) = f (x) dg(t) dt. Further, to meet the boundary conditions (,t) = and (,t) = for < t < by such a product, it is necessary that f () = and f () =. Now, dividing by f (x)g(t), the first equation can be re-written as 1 d 2 f (x) 1 dg(t) f (x) dx 2 = g(t) dt which shows that a function of x is equal to a function of t. The only way that can happen is if both functions are, in fact, the same constant. Let us write the constant as, anticipating that might be positive. Thus 1 d 2 f (x) 1 dg(t) f (x) dx 2 = =, g(t) dt which means that we'll have generated a solution to the pde if f(x) and g(t) satisfy the ode's dg(t) dt + g(t) = and d 2 f (x) dx 2 + f (x) =. These are both linear ode's with constant coefficients. Their general solutions are readily obtained by methods learned in Math 21 or Appl Math 21, and are as follows: g(t) = C exp(t), and f (x) = a exp(i x) + bexp(i x) = Asin( x) + Bcos( x). 2

3 ere C, A, B, a, b are constants; a, b can be related to A, B by using Euler's formula exp(±i) = cos() ± isin(). So far, we have solved the pde in the assumed product form (x, t) = f (x)g(t). Now we demand that the boundary conditions be met. As explained above, that means that f () = and f () =. We see from the general solution for f (x) just given that f () = B and f () = Asin( ) + B cos( ). Thus f () = requires that B=, and then f () = requires that Asin( ) =. The latter equation tells us that either A =, which is of no interest, for then we would have an identically zero solution of the pde, or else that sin( ) =. Thus to have a non-zero solution for which meets both the pde and the boundary conditions, we must choose so that =, 2, 3,... (that is, = n n 2 2 / 2, where n is any positive integer). The corresponding function f (x), associated with = n, may be denoted as f (n) (x) and is given by where A n is an arbitrary constant. f (n) (x) = A n sin(nx / ) An important property of the functions f (n) (x) is their orthogonality, expressed by f (n) (x) f (m) (x)dx = for m n, because sin(n x / )sin(m x / )dx = for m n. You should check the latter result by doing the integral [use sin() = (1 / 2i)(e i e i ) ]. When n=m, you should also find that the result of the integral is just /2. Aside: The equations above controlling the determination of and f (x) constitute a characteristic value problem, also called an eigenvalue problem, for a continuous system. We may say that the n are the characteristic values or eigenvalues and the associated f (n) (x) are the characteristic functions or eigenfunctions. Appendix 2 explores the relation between such continuous eigenvalue problems, associated with homogeneous differential equations and boundary conditions, and the matrix eigenvalue problem that you have studied in Math 21b or Appl Math 21b or an analogous course. It also explains the deeper reasons why the n of our 3

4 problem above are positive and why the functions f (n) (x) are orthogonal. Now, back to the pde problem: At this point we can combine our solutions for f (x), g(t) and to produce an infinite number of product solutions of type (x, t) = D n sin(nx / )exp(n 2 2 t / 2 ) with n = 1, 2, 3, 4,... and with arbitrary constants D n. Each of these forms satisfies the pde and boundary conditions but none satisfy the initial condition that (x,)= init (x), except for fortuitously chosen init (x). We therefore seek to satisfy the initial condition by summing such product forms for all n, trying (x,t) = D n sin(n x / )exp(n 2 2 t / 2 ). n=1 Now, we hope to be able to choose the D n so that we meet the initial condition (x,) D n sin(n x / ) = init (x) for < x <. n=1 Let us assume that this sine series is complete (it is) in the sense that any reasonable function init (x) defined on < x < can be so expressed on that interval. (See the closing paragraph of Appendix 2 for how the sine series relates to the standard Fourier series). Then the orthogonality condition gives us an easy way of obtaining the D n, as follows: We multiply both sides of the last expression by sin(mx / ) and then integrate from to (assuming that interchange of integration and summation is valid for our infinite series, like it would obviously be for any finite series of terms). That gives D n sin(n x / )sin(m x / )dx = init (x)sin(m x / )dx for m = 1, 2, 3, 4,.... n=1 By orthogonality, every integral in the infinite sum vanishes except for that with n=m, and that integral, as we can easily show ourselves, just gives /2. Thus we can solve at once for which completes the solution. D m = (2 / ) init (x)sin(m x / )dx, m = 1, 2, 3, 4,..., Uniform initial : Consider the special case init (x) = o (a constant) on < x <. Then 4

5 the integral giving D m vanishes whenever m is an even integer, and gives D m = 4 o / m when m is an odd integer. Thus the solution for this case is (x,t) = 4 o sin(n x / ) exp(n 2 2 t / 2 ) n n=1,3,5,... That is a rapidly convergent series except for extremely small values of 2 t / 2. The average in the slab may then be computed as (t) 1 (x,t)dx = 8 o 2 n=1,3,5,... 1 n 2 exp(n2 2 t / 2 ) which is also rapidly convergent. Solution for a half-space, with uniform initial, by a similarity method: The half-apace is the limiting case of the slab of infinite thickness,. Note that the above series solution is not useful in that limit. When init (x) = o (a constant), however, there is a simple solution: The problem: Determine = (x,t) which solves the pde 2 x 2 = t in the domain < x < and < t < of the x,t plane subject to the and the initial condition at t=: (x,)= o for < x <, boundary condition at x=: (,t) = for < t <. From the point of view of "dimensional analysis", we may note that this problem specification contains no characteristic length. Thus the only way that x and t could appear in the solution is through the non-dimensional combination x / t. From a more abstract point of view, we find that there is a self-similar transformation which leaves the description of this problem invariant (see homework). We shall therefore seek a solution in the form (x, t) = F() where = (x,t) x 2 t ; 5

6 the 2 is just inserted for later convenience. Thus, direct calculations give spatial derivatives as x = df() d x = 1 2 t and the time derivative as df() d and 2 x 2 = 1 2 t d 2 F() d 2 x = 1 d 2 F() 4t d 2, t = df() d t = 1 x df() 2 2 t3/2 = df() d 2t d ence, by inserting into the pde, we find that it will be satisfied if we satisfy the ode. 1 d 2 F() 4t d 2 = df() 2t d, or d 2 F() d df() d =. Fortunately, this is straightforward to solve (which is not true of most second-order ode's with variable coefficients) because it can be reduced to a first-order ode for df()/d. Let G() = df()/d. Then dg() d so that direct integration gives + 2G() exp( 2 ) d d [exp(2 )G()] = G() = (2 / )Bexp( 2 ), implying df()/d = (2 / )Bexp( 2 ) where B is an arbitrary constant and putting in the the 2/ simplifies the next step. Integrating once again, and noting that the error function erf () is defined by erf() = 2 exp(u 2 )du [note: erf(+) = 1, erf() =, erf() = 1] (a homework helps you to verify that erf(+) = 1 ), we therefore have the solution where A, like B, is an arbitrary constant. (x,t) = F() = A + Berf() = A + Berf(x /2 t ) We must now choose those constants to fit the boundary and initial conditions. To fit the boundary condition (,t) = for < t <, we note that as x at any fixed t on < t <, 6

7 and erf () so that F A. Thus the boundary condition requires that A =. And to fit the initial condition (x,)= o for < x <, we note that as t for any fixed x on < x <, and erf () 1, so that F A + B = B. Thus the initial condition requires that B = o, and therefore the solution of the pde in the half-space is (x,t) = o erf(x /2 t ). Also of some interest is the form of the flux term (flux of heat, or of diffusing species, or of coins, depending on our interpretation), which is given by Q(x,t) = K( x,t)/x, corresponding to Q(x,t) = K[ o / t ]exp(x 2 /4t). ence Q(,t) = K o / t at the boundary of the half-space. That can be interpreted as showing that t is an effective length scale near the boundary over which falls from a value near o to. Appendix 1: Origin of the diffusion pde The diffusion pde begins its life as a pair of equations. Both involve a quantity [= (x,t)] of interest. Different interpretations of are as follows: Solution chemistry: = local mass of solute per unit volume of solution at position x at time t. Thermal physics (heat conduction): = local temperature at position x in a solid body at time t. Currency distribution: = number of coins of a certain type, per unit land area, at place x within some economic region at time t. In all cases, something flows at a rate Q which is negatively proportional to the gradient in the quantity : Q = K x where K > (so the flow tends to go from regions of higher to those of lower. Interpretations of the flux expression are as follows: Solution chemistry: Q = mass of solute molecules crossing unit area in unit time; K = diffusivity; expression for Q is Fick law of diffusion. Thermal physics (heat conduction): Q = rate of energy flow as heat crossing unit area in unit time; K = thermal conductivity; expression for Q is Fourier law of heat conduction. Currency distribution: Q = number of particular type of coin for which home of owner crosses 7

8 unit length along the land surface per unit time; K = coefficient proportional to frequency of economic transactions (transactions cause sellers in a region with an abundance of such coins to lose them to purchasers coming from a region with a relative paucity of such coins). Also, in all cases, whatever is conducted by the flow Q is conserved during that flow (i.e., it is moved from place to place but not created or destroyed). That conserved something has local density proportional to ; i.e., density = C (where the constant coefficient C may be unity in some interpretations). Conservation during the flow is expressed by requiring that Q x = C (figure out why!), t with interpretations as follows: Solution chemistry: C = 1; mass of solute is conserved. Thermal physics (heat conduction): C = specific heat per unit volume of solid, so that C = internal energy of solid per unit volume; energy is conserved. Currency distribution: C = 1; total number of coins in circulation is conserved. Substituting the expression for flow rate Q into the conservation equation then gives (assuming constant K) the diffusion pde: 2 x =, where = K 2 t C. This type of pde also arises in certain simplified problems of the dynamics of motion in viscous fluids, and in describing time-dependent decay of the magnetic field in a stationary electric conductor. Appendix 2: Relation of the continuous and matrix eigenvalue problems You should see an analogy between the solution we obtained above for f (x) and the kind of characteristic value, or eigenvalue, problem that you have studied in Math 21b or Appl Math 21b. We were faced with solving the homogeneous problem Lf (x) = f (x) on < x <, where L = d 2 / dx 2 and the solution was further constrained by the homogeneous conditions f () = f () =. Solutions were found only for certain values of, called the characteristic values, or eigenvalues. That is analogous to the matrix characteristic value problem (Math 21b) for a real square matrix [A], i.e., the problem [ A]{v} = {v}. 8

9 There also, solutions can be found only for certain values of, also called the characteristic values or eigenvalues. Indeed, since our f (x) is defined at an infinite number of points, we may think of our problem as one that corresponds in some sense to an infinite number of elements of {v}; since the matrix problem has as many eigenvalues (allowing for repeated roots) as there are elements of {v}, such makes it plausible that our continuous case has a countably infinite number of eigenvalues. The analogy goes further: Recall that if the matrix problem satisfies {u} [A]{v} = {v } [A]{u} for all {u},{v} (the prime denotes matrix transpose), which is the same as saying that the matrix [A] is symmetric, [ A] = [A], then the following are true: (i) The eigenvalues are real numbers, and (ii) The eigenvectors are orthogonal to one another, at least if they correspond to distinct eigenvalues. The precise statement of (ii) is that if {u(n) }, {u (m) } are a pair of eigenvectors, respectively associated with the pair n, m of eigenvalues, then ( n m ){u (n) } {u (m) } =. In the continuous problem the property analogous to {u} [A]{v} = {v } [A]{u} is that h(x)lf (x)dx = f (x)lh(x)dx for all functions h(x), f (x) meeting the homogeneous boundary conditions f () = f () = and h() = h() =. (Try to prove that our L = d 2 / dx 2 satisfies that condition; an integration by parts will get you there). Operators L which satisfy that condition are said to be self-adjoint. Then, analogously to the matrix case, one may show that (i) The eigenvalues corresponding to such operators are real, and (ii) The eigenfunctions are orthogonal. That is, if f (n) (x), f (m) (x) are a pair of eigenfunctions of a self-adjoint operator L, respectively associated with the pair n, m of its eigenvalues, then ( n m ) f (n) (x) f (m) (x)dx =, as we have explicitly seen above for the f (n) (x) = sin(nx / ) corresponding to our operator L = d 2 / dx 2. This discussion is a foretaste of Sturm-Liouville theory, which we will encounter later in the course. That theory deals with equations of the type Lf (x) = r(x) f (x) where L is a secondorder differential operator and r(x) is a given function. If the operator L is self-adjoint in the 9

10 sense explained above for the class of functions h(x), f (x) meeting suitable homogeneous boundary conditions at the end points (,) of the interval on which the operator is defined, then the eigenfunctions of the operator L, meeting those same boundary conditions, satisfy the orthogonality condition ( n m ) f (n) (x) f (m) (x)r(x)dx =. Such orthogonality conditions, together with the completeness of the set of eigenfunctions, enables generalizations of the Fourier series concept, in that arbitrary functions defined on (,) can be expressed as a series in the eigenfunctions, with series coefficients readily computed because of the orthogonality property. Finally, we should recall that in the matrix case with [ A] = [A], if {v} [A]{v} > for all {v} not identically zero, which means that if [A] is positive definite, then each eigenvalue is not only real but also positive. Similarly, our self-adjoint differential operator L = d 2 / dx 2 has the feature f (x)lf (x)dx > or all continuous functions f (x) meeting f () = f () = but which are not identically zero (try to prove that too; again, an integration by parts will get you there). Self-adjoint operators L meeting that inequality are said to be positive definite operators; such operators have real and positive eigenvalues, as we have already seen in our particular case. If you have studied Fourier series, you understand that a function (x) defined over a portion of the x axis between, say, some point x 1 and x 1 +P should correspond to a Fourier series (if its period is chosen as P) in the form (x) = a + [a n cos(2nx / P) + b n sin(2nx / P)] for x 1 < x < x 1 + P n=1 ence you might be puzzled by the lack of cosine terms in the series for init (x), i.e. the series given in the expression (slightly rearranged here) init (x) = D n sin(n x / ) for < x <. n=1 The latter is a Fourier sine series. It is related to the standard Fourier series in the following way: Let init (x) be defined on < x <. We then define a new function (x) on both < x < and < x < such that (x) = init (x) on < x <, but that (x) is continued as an odd function relative to x=, i.e., defined by (x) = (x) on < x <. If we now identify x 1 with, and P with 2, then the standard Fourier series for (x) has identically zero coefficients for its cosine terms (cosines are even, not odd, relative to x=) and just reduces to the sine series given above for init (x). 1